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Zarrin [17]
3 years ago
14

Plz help me with these 2 problems I really need it and I need to know how you to the answer so I can show my work

Mathematics
1 answer:
ELEN [110]3 years ago
7 0

Answer:

13.) 0.19625    15.) 37.68

Step-by-step explanation:

Formula for area of a circle: A=πr²

13.) 3.14 x .25² = 0.19625

Formula for circumference for a circle: C=πD  or C=2πr

15.) (6 x 2) x 3.14 = 37.68

Hope this helped :  )

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Using the z-distribution, it is found that the 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have that:

  • 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.
  • 49% out of 2001 U.S. adults play video games, hence \pi = 0.49, n = 2001.

The lower bound of the interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 - 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.4681

The upper bound of the interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 + 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.5119

The 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.

To learn more about the z-distribution, you can take a look at brainly.com/question/25730047

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